Question: Graph this system of equations and solve. $y = \dfrac{1}{3} x + 5$ $-7x-3y = 9$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Explanation: The y-intercept for the first equation is $5$ , so the first line must pass through the point $(0, 5)$ The slope for the first equation is $\dfrac{1}{3}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $3$ positions to the right. $3$ positions to the right. Graph the blue line so it passes through $(0, 5)$ and $(3, 6)$ Convert the second equation, $-7x-3y = 9$ , to slope-intercept form. $y = -\dfrac{7}{3} x - 3$ The y-intercept for the second equation is $-3$ , so the second line must pass through the point $(0, -3)$ The slope for the second equation is $-\dfrac{7}{3}$ . Remember that the slope tells you rise over run. So in this case for every $7$ positions you move down (because it's negative) You must also move $3$ positions to the right. $3$ positions to the right. $7$ positions down from $(0, -3)$ is $(3, -10)$ Graph the green line so it passes through $(0, -3)$ and $(3, -10)$ The solution is the point where the two lines intersect. The lines intersect at $(-3, 4)$.